Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $a = \dfrac{18r - 60}{-9} \times \dfrac{5r}{2(3r - 10)} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ (18r - 60) \times 5r } { -9 \times 2(3r - 10) } $ $ a = \dfrac {5r \times 6(3r - 10)} {-9 \times 2(3r - 10)} $ $ a = \dfrac{30r(3r - 10)}{-18(3r - 10)} $ We can cancel the $3r - 10$ so long as $3r - 10 \neq 0$ Therefore $r \neq \dfrac{10}{3}$ $a = \dfrac{30r \cancel{(3r - 10})}{-18 \cancel{(3r - 10)}} = -\dfrac{30r}{18} = -\dfrac{5r}{3} $